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dydlf Group Title

PLEASE HELP MEEE :( consider the points P (-3, 0), Q (a, b) R (3, -3) and S (b+2,a+5). find the coordinates of Q and S if figure PQRS forms a parallelogram. okay so what i did is equate the slope and distances of lines PQ and RS... but i'm still not getting a final answer :(

  • 2 years ago
  • 2 years ago

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  1. dydlf Group Title
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    slope pq= slope rs--> b/(a+3)=(a+8)/(b-1) distance pq = distance rs --> (a+3)^2+b^2=(b-1)^2+(a+8)^2 so b=5a+28 tried to do systems but then there's no answer?!?!?! based on http://www.wolframalpha.com/input/?i=%28a%2B3%29%5E2%2Bb%5E2%3D%28b-1%29%5E2%2B%28a%2B8%29%5E2%3B+b%2F%28a%2B3%29%3D%28a%2B8%29%2F%28b-1%29 huhuhu someone help please :((

    • 2 years ago
  2. ghass1978 Group Title
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    did you try solving it with vectors?

    • 2 years ago
  3. amistre64 Group Title
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    im thinking of midpoints :)

    • 2 years ago
  4. amistre64 Group Title
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    (P+R)/2 = (Q+S)/2 P+R = Q+S

    • 2 years ago
  5. amistre64 Group Title
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    but that just gets me parallel lines that have no solution :/ a+b = -2 a+b = -8

    • 2 years ago
  6. amistre64 Group Title
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    is there a figure to pull information off of?

    • 2 years ago
  7. amistre64 Group Title
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    |dw:1345725961534:dw| we need to develop a notion of the rule that defines Q and S to narrow down the options eh

    • 2 years ago
  8. dydlf Group Title
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    what did you mean by your last message, amistre64? btw, is there anything wrong with my solution?

    • 2 years ago
  9. amistre64 Group Title
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    a parallelagram is defined as opposite sides are parallel; i was just looking for "orbits" that would define the possibilities of 2 other points.

    • 2 years ago
  10. dydlf Group Title
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    ghass1978, nope i don't remember how to use vectors... it's not part of our exam. sorry

    • 2 years ago
  11. amistre64 Group Title
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    your initial ideas look good in that they are defining valid properties

    • 2 years ago
  12. amistre64 Group Title
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    is there a diagram that goes with this? or is it just a wordy problem?

    • 2 years ago
  13. dydlf Group Title
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    just that, exactly that :( waaaaahhh :(

    • 2 years ago
  14. dydlf Group Title
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    but it says PQRS which means the sequence of the coordinates would be that way. as in Q "between" P and R... if that's what you're wondering

    • 2 years ago
  15. amistre64 Group Title
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    |dw:1345726700410:dw| without using the relationship between Q and S, we can get any number of values for Q and S

    • 2 years ago
  16. amistre64 Group Title
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    Q(0,0) S(2,5) if we adjust this so that the midpoint is at 0,-1.5 it satisfies the rule Q(-1,-4) S(1,1) is a solution

    • 2 years ago
  17. dydlf Group Title
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    wait wait i don't get your drawing??? how did you get those? :( sorry!! i don't think we were taught that method yet?

    • 2 years ago
  18. dydlf Group Title
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    why get the midpoint?

    • 2 years ago
  19. amistre64 Group Title
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    draw a parallelgram|dw:1345727410942:dw|

    • 2 years ago
  20. amistre64 Group Title
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    |dw:1345727435579:dw||dw:1345727448036:dw|

    • 2 years ago
  21. amistre64 Group Title
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    the midpoint of PR is the midpoint of QS , define some points that match the rule Q(a,b) S(b+2,a+5) and resolve the midpoint by moving the points about

    • 2 years ago
  22. amistre64 Group Title
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    any scaled values of Q and S also work if there is a given picture to conform to, we can narrow down the odds of what they want tho

    • 2 years ago
  23. amistre64 Group Title
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    |dw:1345727657560:dw|

    • 2 years ago
  24. SWAG Group Title
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    Nice work @amistre64

    • 2 years ago
  25. amistre64 Group Title
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    thnx :)

    • 2 years ago
  26. dydlf Group Title
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    the right answer is S(5, -4) and Q(-9,3) ?? just found out now... how do you get that then? btw i will close this, have another question to ask.

    • 2 years ago
  27. amistre64 Group Title
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    -9, 3 5,-4 ------ -4,-1 ; midpoint = -2,-1/2 -3, 0 3, -3 ------ 0,-3 ; midpoint = 0, -3/2 you either have some misinformation in the post, or the answer is wrong.

    • 2 years ago
  28. dydlf Group Title
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    okay wait so i did the midpoint thing but i'm still lost... midpoint of pr is (0, -3/2) midpoint of qs is ([a+b+2]/2, [a+b+5]/2) i tried equating [a+b+2]/2=0 and [a+b+5]/2=-3/2 then i got... a+b=-2 a+b=-8 ??? so how did you get Q(-1,-4) S(1,1) ?

    • 2 years ago
  29. dydlf Group Title
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    nevermind, i believe our teacher gave the wrong given. i believe s should have been (a+2, b+5)

    • 2 years ago
  30. amistre64 Group Title
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    I essentially created a set of points in a different place that matched the given rule. Then I moved those points into the proper position.

    • 2 years ago
  31. amistre64 Group Title
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    using your new rule for Q and S .... Q(a,b) S(a+2, b+5) select any values you want for a and b; say a=6 and b=-1 .... random picks Q(6,-1) S(6+2, -1+5) Q(6,-1) S(8, 4) now that I have a set of points that matches the rule, lets determine its own midpoint as is (7 , 3/2) now i want this midpoint to be at (0,-3/2), so lets see how we adjust it to get there (7 , 3/2) -7, -6/2 <---- i need to move Q and S by the same -------- (0, -3/2) since we have to move the midpoint by (-7,-3), we will have to move Q and S by the same amount Q(6,-1) S(8, 4) -7 -3 -7 -3 ------- -------- Q(-1,-4) S(1, 1) now we have a set of points Q and S that match the given rule; and is located in the proper spot

    • 2 years ago
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